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Right here we’ll be taught subtracting 2digit numbers with
borrowing. The subtractions with borrowing are solved stepbystep in 4
alternative ways.
When we have to subtract larger quantity from a smaller quantity in ones place, we regroup tens into ones.
Regrouping
We will write a 2digit quantity in numerous methods by regrouping tens and ones.
16 will be written as
23 will be written as
Allow us to perceive this with the assistance of an instance.
A shopkeeper sells bananas. He has 6 bunches of 10’s and 5 single bananas.
Nairitee needs to purchase 37 bananas from the shopkeeper. The shopkeeper provides her 3 bunches of 10’s and breaks one bunch of 10’s to present 7 bananas.
The shopkeeper is now left with what number of bananas?
Column Methodology
Prepare the numbers in columns and subtract. In ones place, 5 is smaller than 7. We can not take away 7 from 5. So, we break one ten into 10 ones. Now we’ve, 15 ones in ones place and 5 tens in tens place.
15 ones – 7 ones = 8 We write 8 in ones place. Subtract 3 tens from 5 tens. 5 tens – 3 tens = 2 tens Write 2 in tens place. 
So, 65 – 37 = 28
Laboredout examples on subtracting 2digit numbers with borrowing:
1. Subtract 9 from 15.
Resolution:
T O
1 5
– 9
Since, 5 < 9, so 9 can’t be subtracted from 5. So, 1 ten, i.e., 10 ones is borrowed from the digit 1 of tens place. Now one ten, i.e., 10 ones are added to five ones to make it 15 ones. Now 15 ones – 9 ones = 6 ones.
Subsequently, 15 – 9 = 6
2. Subtract 37
from 65
Resolution:
The numbers are positioned in column type, with the smaller
quantity 37 written beneath the better quantity 65.
T O
1 T → 10
6 5
– 3
7
2 8
(i) first ones are subtracted as 5 < 7 or 7 > 5. So, 7
can’t be subtracted from 5.
(ii) Now 1 ten is borrowed from 6 tens leaving 5 tens there.
(iii) 1 ten = 10 ones. So, 10 ones are added to five ones
making the sum 15 ones
(iv) 7 ones are subtracted from 15 ones i.e., 15 ones – 7
ones = 8 ones. This 8 is written in a single’s column.
(v) Now tens are subtracted. At ten’s place there are 5 tens
left. So 5 tens – 3 tens = 2 tens. So, 2 is written in ten’s column.
(vi) Subsequently, 65 – 37 = 28.
3. Subtract 28
from 83
Resolution:
The smaller quantity 28 is written beneath better quantity 83 in
column type and ones are subtracted first, then the tens.
T O
1 T → 10
8 3
– 2
8
5 5
(i) 3 < 8, so 1 ten, i.e., 10 ones are borrowed from 8
tens with 7tens remaining there.
(ii) Now, 1 ten + 3 = 10 + 3 = 13 ones. So, 13 ones – 8 ones
= 5 ones.
(iii) 7 tens – 2 tens = 5 tens.
Subsequently, 83 – 28 = 55
4. Subtract 69
from 92
Resolution:
The smaller quantity 69 is written beneath better quantity 92 in
column type and ones are subtracted first, then the tens.
T O
1 T → 10
9 2
– 6
9
2 3
(i) 10 + 2 = 12; 12 O
– 9 O = 3 O
(ii) 8 T – 6 T = 2 T
Subsequently, 92 – 69 = 23
Subtraction with Borrowing:
5. Allow us to subtract 5 from 23.
Step I: Prepare the numbers into tens and ones. 

Step II: We can not subtract 5 from 3. So, borrow 1 ten from the tens column. 1 ten = 10 ones. Take these 10 ones to those column. This offers: 10 + 3 = 13 ones 13 – 5 = 8 

Step III: Subtract the tens. Now since we borrowed 1 ten, we’re left with 1 ten within the tens column. 
Thus, 23 – 5 = 18
6. Allow us to subtract 37 from 53.
Step I: Prepare the numbers into tens and ones. 

Step II: We can not subtract 7 from 3. So, borrow 1 ten from the tens column. 1 ten = 10 ones. Take these 10 ones to those column. This offers: 10 + 3 = 13 ones 13 – 7 = 6 

Step III: Since we’ve borrowed 1 ten from 5 tens, we’re left with 4 tens. Now subtract 3 tens from 4 tens. 4 – 3 = 1 
Thus, 53 – 37 = 16
Questions and Solutions on Subtracting 2Digit Numbers with Borrowing (Regrouping):
1. Regrouping the tens and ones.
(i) 
24 = 
2 
tens + 
4 
ones = 
1 
ten + 
14 
ones 
(ii) 
57 = 
___ 
tens + 
___ 
ones = 
4 
tens + 
___ 
ones 
(iii) 
64 = 
___ 
tens + 
___ 
ones = 
___ 
tens + 
14 
ones 
(iv) 
48 = 
___ 
tens + 
___ 
ones = 
3 
tens + 
___ 
ones 
2. Subtracting 1digit quantity from 2digit quantity with regrouping.
(i) 38 – 9 = _____
(ii) 62 – 7 = _____
(iii) 44 – 6 = _____
(iv) 67 – 8 = _____
(v) 75 – 7 = _____
(vi) 94 – 8 = _____
(vii) 74 – 5 = _____
(viii) 51 – 2 = _____
(ix) 95 – 6 = _____
(x) 42 – 3 = _____
(xi) 53 – 4 = _____
(xii) 62 – 4 = _____
(xiii) 67 – 8 = _____
(xiv) 32 – 5 = _____
(xv) 22 – 9 = _____
(xvi) 93 – 3 = _____
Solutions:
2. (i) 29
(ii) 55
(iii) 38
(iv) 59
(v) 68
(vi) 86
(vii) 69
(viii) 49
(ix) 89
(x) 39
(xi) 49
(xii) 58
(xiii) 59
(xiv) 27
(xv) 13
(xvi) 90
3. Subtracting 2digit quantity with regrouping.
(i) 80 – 17 = _____
(ii) 38 – 29 = _____
(iii) 71 – 34 = _____
(iv) 47 – 19 = _____
(v) 86 – 27 = _____
(vi) 65 – 47 = _____
(vii) 51 – 25 = _____
(viii) 62 – 37 = _____
(ix) 73 – 57 = _____
(x) 94 – 46 = _____
(xi) 46 – 28 = _____
(xii) 56 – 27 = _____
(xiii) 57 – 19 = _____
(xiv) 31 – 24 = _____
(xv) 41 – 16 = _____
(xvi) 53 – 29 = _____
Solutions:
3. (i) 63
(ii) 9
(iii) 37
(iv) 28
(v) 59
(vi) 18
(vii) 26
(viii) 25
(ix) 16
(x) 48
(xi) 18
(xii) 29
(xiii) 38
(xiv) 7
(xv) 25
(xvi) 24
4. Subtract.
(i) 
– 
T O 6 3 4 6 __________ 
(ii) 
– 
T O 3 4 2 9 __________ 
(iii) 
– 
T O 9 3 5 8 __________ 
(iv) 
– 
T O 7 6 5 9 __________ 
(v) 
– 
T O 3 2 1 5 __________ 
(vi) 
– 
T O 4 4 3 6 __________ 
(vii) 
– 
T O 5 2 2 8 __________ 
(viii) 
– 
T O 3 2 1 9 __________ 
(ix) 
– 
T O 8 3 6 9 __________ 
(x) 
– 
T O 7 5 2 6 __________ 
Solutions:
4. (i) 17
(ii) 5
(iii) 35
(iv) 17
(v) 17
(vi) 8
(vii) 24
(viii) 13
(ix) 14
(x) 49
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